Question

The sides of a right angled triangle are integers. The length of one of the sides is $12$. The largest possible radius of the incircle of such a triangle is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (D)

In right angle $\Delta ABC$
Let $AB=12$
$BC=x+r$
$AC=12+x-r$

$A{C}^2=A{B}^2+B{C}^2$
$\Rightarrow (12+x-r{)}^2=(12{)}^2+(x+r{)}^2$
$\Rightarrow 144+24(x-r)+(x-r{)}^2$
$=144+(x+r{)}^2$
$\Rightarrow 24(x-r)=(x+r{)}^2-(x-r{)}^2$
$\Rightarrow 24(x-r)=4xr$
$\Rightarrow 6(x-r)=xr$
$\Rightarrow r=\frac{6x}{6+x}$
At $x=6,r=3$
At $x=12,r=4$
At $x=30,r=5$
$\therefore$ Maximum radius $=5$